# Inner product with maximally entangled state

1. Dec 14, 2011

### rmp251

Consider the maximum inner product of a density operator with a maximally entangled state. (So, given a density operator, we're maximizing over all maximally entangled states.)

I'm pretty sure the minimum value (a lower bound on the maximum) for this is 1/n2, using the density operator 1/n2 identity. How can I prove that is the minimum?

Thanks!
Reuben

2. Dec 15, 2011

### tom.stoer

can you please write down the density operator you are talking about; and what you mean by 'maximum inner product of a density operator with a maximally entangled state'.

3. Dec 15, 2011

### rmp251

Sorry I realize that was a little incomplete. Let $\mathcal{X}$ and $\mathcal{Y}$ be complex Euclidean spaces with dim($\mathcal{X}$)=dim($\mathcal{Y}$)=n.

Define

$M(\rho) = \max\left\{<u u^{\ast},\rho>\,:\,u\in\mathcal{X}\otimes\mathcal{Y}\;\text{is maximally entangled} \right\}$

for any density operator $\rho$ on $\mathcal{X}\otimes\mathcal{Y}$.

For $\rho=\frac{1}{n^2}\text{I}$ we get $M=\frac{1}{n^2}$

How can I show that that is the minimum value of $M$ over all possible density operators $\rho$?